BH2- Lewis structure

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BH₂^– has one boron atom and two hydrogen atoms.

In BH₂^– Lewis structure, there are two single bonds around the boron atom, with two hydrogen atoms attached to it, and on the boron atom, there is one lone pair.

Also, there is a negative (-1) charge on the boron atom.

Steps

Here’s how you can easily draw the BH₂^– Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, boron lies in group 13, and hydrogen lies in group 1.

Hence, boron has three valence electrons and hydrogen has one valence electron.

Since BH₂^– has one boron atom and two hydrogen atoms, so…

Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of two hydrogen atoms = 1 × 2 = 2

Now the BH₂^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 3 + 2 + 1 = 6

Learn how to find: Boron valence electrons and Hydrogen valence electrons

• Second, find the total electron pairs

We have a total of 6 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 6 ÷ 2 = 3

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Hence, here we have to assume that the central atom is boron.

Therefore, place boron in the center and hydrogens on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 3 electron pairs. And two B — H bonds are already marked. So we have to only mark the remaining one electron pair as a lone pair on the sketch.

Also remember that boron is a period 2 element, so it can not keep more than 8 electrons in its last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for boron, there is one lone pair.

Mark the lone pair on the sketch as follows:

#3 If needed, mention formal charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For boron atom, formal charge = 3 – 2 – ½ (4) = -1

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

Here, the boron atom has a charge, so mark it on the sketch as follows:

In the above structure, you can see that the central atom (boron) doesn’t form an octet. But, boron has an exception that it does not require eight electrons to form an octet. So no need to worry about the octet rule here.

Now there is still a negative (-1) charge on the boron atom.

This is okay, because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is boron.

Therefore, this structure is the most stable Lewis structure of BH₂^–.

And since the BH₂^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: IO₃^– Lewis structure

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External links

• https://lambdageeks.com/bh2-lewis-structure/
• https://www.pinterest.com/pin/bh2-lewis-structure-in-2021–661395895284903532/
• https://www.bartleby.com/questions-and-answers/the-lewis-diagram-forbh2-is-theelectron-pairgeometry-around-thebatom-inbh2-is-______-there-are-_____/8f85d8e8-22e8-4ee7-bf47-a9c592bd3f2d
• https://brainly.com/question/4083242